Check if $ x-5 $ is a factor of $ x^{3}-2x^{2}-23x-60 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&1&-2&-23&-60\\& & 5& 15& \color{black}{-40} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-8}&\color{orangered}{-100} \end{array} $$Because the remainder $ \left( \color{red}{ -100 } \right) $ is not zero, we conclude that the $ x-5 $ is not a factor of $ x^{3}-2x^{2}-23x-60$.
Explanation
First we need to create a synthetic division table.
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-2&-23&-60\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 1 }&-2&-23&-60\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-2&-23&-60\\& & \color{blue}{5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 5 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}5&1&\color{orangered}{ -2 }&-23&-60\\& & \color{orangered}{5} & & \\ \hline &1&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-2&-23&-60\\& & 5& \color{blue}{15} & \\ \hline &1&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 15 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}5&1&-2&\color{orangered}{ -23 }&-60\\& & 5& \color{orangered}{15} & \\ \hline &1&3&\color{orangered}{-8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-2&-23&-60\\& & 5& 15& \color{blue}{-40} \\ \hline &1&3&\color{blue}{-8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -60 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ -100 } $
$$ \begin{array}{c|rrrr}5&1&-2&-23&\color{orangered}{ -60 }\\& & 5& 15& \color{orangered}{-40} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-8}&\color{orangered}{-100} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -100 }\right)$.